Light of wavelength \(\lambda\) strikes a photoelectric surface and electrons are ejected with energy \(E\). If \(E\) is to be increased to twice the original value, the wavelength changes to \(\lambda_1\)
Show Hint
Due to work function, wavelength is always less than half for doubling energy.
Step 1: Understanding the Concept:
Einstein's photoelectric equation relates the maximum kinetic energy of ejected electrons to the frequency (or wavelength) of incident light and the work function of the metal. Step 2: Key Formula or Approach:
Photoelectric equation: \(K_{\text{max}} = \frac{hc}{\lambda} - \phi\)
Given initial kinetic energy \(K_1 = E\) and final kinetic energy \(K_2 = 2E\). Step 3: Detailed Explanation:
Write the equations for both cases:
Case 1: \(E = \frac{hc}{\lambda} - \phi \implies \frac{hc}{\lambda} = E + \phi\)
Case 2: \(2E = \frac{hc}{\lambda_1} - \phi \implies \frac{hc}{\lambda_1} = 2E + \phi\)
We want to compare \(\lambda_1\) and \(\lambda/2\). Let's calculate \(2 \cdot (hc/\lambda)\):
\[ \frac{2hc}{\lambda} = 2(E + \phi) = 2E + 2\phi \]
Now compare this with \(hc/\lambda_1\):
\[ \frac{hc}{\lambda_1} = 2E + \phi \]
Clearly, \(\frac{hc}{\lambda_1}<\frac{2hc}{\lambda}\) since \(\phi\) is positive.
Divide by \(hc\):
\[ \frac{1}{\lambda_1}<\frac{2}{\lambda} \implies \frac{1}{\lambda_1}<\frac{1}{\lambda/2} \]
Since both \(\lambda_1\) and \(\lambda/2\) are positive, taking the reciprocal reverses the inequality:
\[ \lambda_1>\lambda/2 \]
Step 4: Final Answer:
The new wavelength \(\lambda_1\) must be greater than \(\lambda/2\).